Topology: Subset of A Continuous Function

[darkchild]

Homework Statement

Let $$f$$ be a real-valued function defined and continuous on the set of real numbers. Which of the following must be true of the set $$S={f(c):0<c<1}?$$

I. S is a connected subset of the real numbers.
II. S is an open subset of the real numbers.
III. S is a bounded subset of the real numbers.

Homework Equations

none

The Attempt at a Solution

The answer is I and III only. I'm confused because I thought that I. implied II. Isn't a connected set necessarily open?

 

[Kreizhn]

Why would a connected set necessarily be open? [0,1] is connected but not open.

 

[micromass]

Not at all! Connectedness does not imply open. A simple counterexample is [0,1], this is connected but not open.

 

[darkchild]

Hmm...well, my textbook defines a subset of a topological space as connected if it's simultaneously relatively open and relatively closed. I thought this applied to S since it is a subset of the real numbers and the real numbers could be a topological space if you include a distance formula...

Do sets and topological spaces have different criteria for being open/closed/etc.?

 

[micromass]

How does your textbook define relatively open/closed?

 

[Kreizhn]

Hmm...well, my textbook defines a subset of a topological space as connected if it's simultaneously relatively open and relatively closed. I thought this applied to S since it is a subset of the real numbers and the real numbers could be a topological space if you include a distance formula...

What your book likely says is that a topological space is connected if and only if there are no non-trivial clopen subsets. You are considering S as a subset of $\mathbb R$ not as a topological space itself.

 

[Kreizhn]

Also, $\mathbb R$ can be made a topological space regardless of whether there is a metric on it or not. The inclusion of metric imposes a particular topology (the one defined by the metric) but one does not need a distance function to define open sets. Indeed, there are some very curious topologies on $\mathbb R$ that are not induced via metrics.

 

[darkchild]

How does your textbook define relatively open/closed?

It says that an open subset of a subspace of a topological space is relatively open.

What your book likely says is that a topological space is connected if and only if there are no non-trivial clopen subsets. You are considering S as a subset of LaTeX Code: \\mathbb R not as a topological space itself.

There's a definition of connectedness for both topological spaces and subsets of topological spaces. Both definitions state: A topological space (or a subset of one) is connected if the only two subsets of it that are both open (relatively) and closed (relatively) are the topological space (or subset) itself and the null set.

Also, LaTeX Code: \\mathbb R can be made a topological space regardless of whether there is a metric on it or not.

Right, I had metric spaces and topological spaces mixed up in my mind when I wrote that.

 

[Kreizhn]

Right. It may not be obvious but your books definition of connected is the same as mine. Consider the classical example of the set
$$[0,1] \cup [2,3]$$
This set is not connected, right? It's pretty obvious when you look at it as being the union of disjoint sets, but let's take a look at it from a clopen (closed and open) point of view.

Consider $X =[0,1] \cup [2,3]$ as a topological space itself, under the subspace topology it inherits from $\mathbb R$. In particular, our normal idea of closed and open intervals being closed and open sets is still true because it's true in $\mathbb R$. But also notice that [0,1] is both closed and open in X (though it's not both closed and open in $\mathbb R$). Why is this true? Well, [0,1] is closed in $\mathbb R$ so it's closed in X as a subspace of $\mathbb R$. Additionally,
$$X\setminus_{[0,1]} = [2,3]$$
That is, its relative complement in X is closed. By definition, a set is open if its complement is closed and so [0,1] is also open!

Thus [0,1] is both open and closed, and is a proper subset of X. So X has a proper, non-trivial clopen subset and hence is not connected.

Edit: The thing to take away from this is that [0,1] is both open and closed as a subset of X, but NOT as a subset of $\mathbb R$!